Question

Find the general solution of the given differential equation. $$ 2 y^{\prime \prime \prime}-4 y^{\prime \prime}-2 y^{\prime}+4 y=0 $$

Short answer

Question: Determine the general solution of the following third-order linear homogeneous differential equation: $$ 2y^{\prime \prime \prime}-4y^{\prime \prime}-2y^{\prime}+4y=0 $$ Answer: The general solution of the given differential equation is: $$ y(x) = C_1 e^{1x} + C_2 e^{\frac{1}{2}x} \cos(\sqrt{7} \frac{x}{2}) + C_3 e^{\frac{1}{2}x} \sin(\sqrt{7} \frac{x}{2}) $$ Where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.

Step-by-step solution

Write down the given differential equation

The given differential equation is: $$ 2y^{\prime \prime \prime}-4y^{\prime \prime}-2y^{\prime}+4y=0 $$

Find the characteristic equation

Replace each derivative by a power of r: $$ 2r^3-4r^2-2r+4=0 $$

Solve the characteristic equation

First, let's factor out 2 from each term: $$ r^3 - 2r^2 - r + 2=0 $$ Now we will try to find its roots. Since it's a cubic polynomial, we can use either trial and error method or cubic formula. Let's try the trial and error method. It turns out that r = 1 is a root, plugging it to the equation, we get: $$ (1)^3 - 2(1)^2 - (1) + 2 = 0 $$ So, r = 1 is indeed a root. Now, we can do synthetic division to find the other factors: $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{r^2} & -r & +2 \\ \cline{2-5} (r-1) & r^3 & -2r^2 & -r & +2 \\ \cline{2-2} & r^3 & -r^2 \\ \cline{2-3} & & -r^2 & -r \\ \cline{3-4} & & & -r & + 1 \\ \cline{4-5} & & & & +3 \end{array} $$ The other factor is \(r^2 - r + 2\). Hence, the characteristic equation is: $$ (r-1)(r^2 - r + 2)=0 $$

Find the roots of the quadratic factor

We have already found one root, r = 1. Now, let's find the roots of the quadratic factor \((r^2 - r + 2)\) using the quadratic formula: $$ r = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ Substitute the values a = 1, b = -1, and c = 2, we get: $$ r = \frac{1\pm\sqrt{(-1)^2-4(1)(2)}}{2(1)} $$ This leads to complex roots: $$ r = \frac{1 \pm i\sqrt{7}}{2} $$

Write the general solution

Since we have one real root and a pair of complex conjugate roots, the general solution of the given differential equation is: $$ y(x) = C_1 e^{1x} + C_2 e^{\frac{1}{2}x} \cos(\sqrt{7} \frac{x}{2}) + C_3 e^{\frac{1}{2}x} \sin(\sqrt{7} \frac{x}{2}) $$ Where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.

Key concepts

Characteristic Equation

In the study of differential equations, the characteristic equation is a crucial concept. It's an algebraic equation obtained from a linear differential equation by replacing each derivative of the function with a power of a variable, typically denoted as 'r'. The purpose of the characteristic equation is to translate the problem of solving a differential equation into the problem of finding the roots of an algebraic equation.
The characteristic equation for the given differential equation, 2y'''-4y''-2y'+4y=0, is derived by replacing derivatives with powers of r, leading to the algebraic equation 2r³-4r²-2r+4=0. This equation captures the essence of the differential equation's behavior. By solving for the roots of the characteristic equation, you can directly assess the nature of the solution to the original differential equation.
Finding the roots can involve a few different techniques: Factoring, using the Rational Root Theorem, or applying synthetic division with trial and error. In the exercise, factoring out 2 leads to a simpler cubic polynomial. The difference between a characteristic equation and regular algebraic equations is primarily in their application; while both are solved using similar methods, the roots of a characteristic equation inform us about the exponential terms in the solution to a differential equation.

Cubic Polynomial

Understanding Cubic Polynomials

A cubic polynomial is an algebraic expression of degree three. It can be represented in the standard form ax³+bx²+cx+d=0, where a, b, c, and d are constants, and a ≠ 0. Solving cubic equations can be more challenging than solving linear or quadratic equations, as they may have up to three real roots or one real root and two complex conjugate roots. Methods to find these roots vary from simple factoring, graphical methods, or more advanced techniques such as Cardano's formula.
For the exercise, the cubic polynomial r³-2r²-r+2=0 was simplified from the characteristic equation. The solution involves factoring to reduce the polynomial into a linear factor and a quadratic factor, which can then be individually solved. Identifying at least one root, typically through the trial and error method or inspection, allows the use of synthetic division to break the cubic polynomial down further, simplifying the process of finding all roots.

Complex Roots

When solving polynomials, complex roots often come into play, especially when dealing with higher-degree polynomials. A complex root occurs when no real number solution exists for a polynomial equation. These roots are in the form a ± bi, where a and b are real numbers, and i is the imaginary unit representing the square root of -1.
In the context of the given differential equation, after determining the real root, we got the quadratic polynomial r² - r + 2. Using the quadratic formula reveals that the remaining roots are complex because the discriminant (b² - 4ac) is negative. Hence, we end up with a pair of complex conjugate roots, which appear in the solution to the differential equation as terms involving exponential, sine, and cosine functions with the imaginary part affecting the frequency of the oscillation. These complex roots are critical in forming a complete solution to differential equations, offering insight into the oscillatory behavior of solutions when the differential equation describes a physical system.